Answer
\[
\boxed{
h_k
= \frac{\,2^{k+1} + (-1)^k\,}{3}.
}
\]
Work Step by Step
We want to solve the recurrence
\[
h_k \;=\; 2^k \;-\; h_{k-1},
\quad\text{for } k\ge1,
\quad\text{with } h_{0}=1.
\]
A convenient first step is to rewrite it as
\[
h_k \;+\; h_{k-1} \;=\; 2^k.
\]
However, it is often more illuminating to “unroll” the given form:
\[
h_k \;=\; 2^k \;-\; h_{k-1}.
\]
Apply it again to \(h_{k-1}\):
\[
h_{k-1} \;=\; 2^{k-1} \;-\; h_{k-2},
\]
so
\[
h_k
= 2^k \;-\;\bigl(2^{k-1} - h_{k-2}\bigr)
= 2^k \;-\; 2^{k-1} \;+\; h_{k-2}
= 2^{k-1} \;+\; h_{k-2}.
\]
In other words, \(h_k\) satisfies the *second*-order recurrence
\[
h_k \;=\; 2^{k-1} \;+\; h_{k-2}.
\]
---
## Computing the first few terms
Let’s list out the terms to see a pattern:
\(h_0 = 1.\)
\(h_1 = 2^1 - h_0 = 2 - 1 = 1.\)
\(h_2 = 2^2 - h_1 = 4 - 1 = 3.\)
\(h_3 = 2^3 - h_2 = 8 - 3 = 5.\)
\(h_4 = 2^4 - h_3 = 16 - 5 = 11.\)
\(h_5 = 2^5 - h_4 = 32 - 11 = 21.\)
\(h_6 = 2^6 - h_5 = 64 - 21 = 43.\)
So the sequence begins
\[
1,\; 1,\; 3,\; 5,\; 11,\; 21,\; 43,\;\dots
\]
---
## Finding a closed form
From the second-order form
\[
h_k = 2^{k-1} + h_{k-2},
\]
we can “unroll” in steps of 2:
- If \(k\) is even, say \(k = 2n\), then
\[
h_{2n}
= 2^{2n-1} + h_{2n-2}
= 2^{2n-1} + 2^{2n-3} + h_{2n-4}
= \dots
= 2^{2n-1} + 2^{2n-3} + \cdots + 2^{1} \;+\; h_{0}.
\]
Since \(h_{0} = 1\), the sum of odd powers
\(\,2^1 + 2^3 + \dots + 2^{2n-1}\) is a known geometric series. One finds
\[
h_{2n}
= \bigl(2^1 + 2^3 + \cdots + 2^{2n-1}\bigr) + 1
= \frac{2^{2n+1} + 1}{3}.
\]
- If \(k\) is odd, say \(k = 2n + 1\), then
\[
h_{2n+1}
= 2^{2n} + h_{2n-1}
= 2^{2n} + 2^{2n-2} + \cdots + 2^{2} \;+\; h_{1}.
\]
Since \(h_{1} = 1,\) the sum of even powers \(2^2 + 2^4 + \dots + 2^{2n}\) can again be summed as a geometric series, yielding
\[
h_{2n+1}
= \frac{2^{2n+2} - 1}{3}.
\]
### A unified single-form expression
Notice we can combine these two cases by introducing \((-1)^k\). Indeed, one finds:
- For even \(k\), \((-1)^k = +1,\) and \(h_k = \tfrac{2^{k+1} + 1}{3}.\)
- For odd \(k\), \((-1)^k = -1,\) and \(h_k = \tfrac{2^{k+1} - 1}{3}.\)
One can check it satisfies the recurrence and the initial condition \(h_0 = 1\).
